Discussion
If \(y = \sec(\tan^{-1}x)\) ,then \(\frac{dy}{dx}\) at \(x = \sqrt{3}\) is equal to
Answer: D
\(\frac{dy}{dx} =\) \(\frac{\sqrt{3}}{2}\).
If \(\theta = \tan^{-1}x\), then \(\sec\theta = \sqrt{1 + x^{2}}\), so \(y = \sqrt{1 + x^{2}}\) and
\[ \begin{aligned} \frac{dy}{dx} &= \frac{x}{\sqrt{1 + x^{2}}} \\ &= \frac{\sqrt{3}}{\sqrt{4}} \\ &= \frac{\sqrt{3}}{2} \end{aligned} \]
at \(x = \sqrt{3}\). \(\frac{1}{2}\) is \(\frac{1}{\sqrt{1 + x^{2}}}\), and 2 is the value of \(y\) itself; the other options do not follow.
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